# How to apply a Gaussian radial basis function kernel PCA to nonlinear data?

I have an assignment to implement a Gaussian radial basis function-kernel principal component analysis (RBF-kernel PCA) and have some challenges here. It would be great if someone could point me to the right direction because I am obviously doing something wrong here.

So, when I understand correctly, the RBF kernel is implemented like this:

$$K(\mathbf{x}_i, \mathbf{x}_j) = \mathrm{exp}\left(- \gamma \|\mathbf{x}_i - \mathbf{x}_j\|^{2}_{2} \right)=\mathrm{exp}\left(- \frac{\|\mathbf{x}_i - \mathbf{x}_j\|^{2}_{2}}{2\sigma^2} \right),$$

where $$\|\mathbf{x}_i - \mathbf{x}_j\|^{2}_{2} = \sum_j(x_{ik} - x_{jk})^2$$ is the squared Euclidean distance between two data points, $$\mathbf{x}_i$$ and $$\mathbf{x}_j$$, and $$\gamma$$ is a free parameter $$\gamma = \frac{1}{2\sigma^2}$$. The $$\sigma^2$$ can be chosen as the variance of the Euclidean distances between all pairs of data points.

To compare my approach to scikit-learn's implementation, I created a simple nonlinear dataset:

### Example dataset

import matplotlib.pyplot as plt

from sklearn.datasets import make_moons
X, y = make_moons(n_samples=100, random_state=123)

plt.figure(figsize=(8,6))

plt.scatter(X[y==0, 0], X[y==0, 1], color='red')
plt.scatter(X[y==1, 0], X[y==1, 1], color='blue')

plt.title('A nonlinear 2Ddataset')
plt.ylabel('y coordinate')
plt.xlabel('x coordinate') ### scikit-learn RBF Kernel PCA

When I used the scikit-learn implementation for dimensionality reduction onto 1 component axis, the classes separate quite nicely.

scikit_kpca = KernelPCA(n_components=1, kernel='rbf', gamma=15)
X_skernpca = scikit_kpca.fit_transform(X)

plt.figure(figsize=(8,6))
plt.scatter(X_skernpca[y==0, 0], np.zeros((50,1)), color='red', alpha=0.5)
plt.scatter(X_skernpca[y==1, 0], np.zeros((50,1)), color='blue', alpha=0.5)

plt.title('First component after RBF Kernel PCA')
plt.show() ### My approach

Somehow, I am not able to reproduce those results. From what I understand, I have to compute all pairwise distances in order to compute the kernel. Then center the Kernel and extract the eigenvector that corresponds to the largest eigenvalue. This is what I have done so far:

from sklearn.preprocessing import KernelCenterer
from scipy.spatial.distance import pdist, squareform
from scipy import exp

# pdist to calculate the squared Euclidean distances for every pair of points
# in the 100x2 dimensional dataset.
sq_dists = pdist(X, 'sqeuclidean')

# Variance of the Euclidean distance between all pairs of data points.
variance = np.var(sq_dists)

# squareform to converts the pairwise distances into a symmetric 100x100 matrix
mat_sq_dists = squareform(sq_dists)

# set the gamma parameter equal to the one I used in scikit-learn KernelPCA
gamma = 15

# Compute the 100x100 kernel matrix
K = exp(gamma * mat_sq_dists)

# Center the kernel matrix
kern_cent = KernelCenterer()
K = kern_cent.fit_transform(K)

# Get the eigenvector with largest eigenvalue
eigvals, eigvecs = np.linalg.eig(K)
eigvals, eigvecs = zip(*sorted(zip(eigvals, eigvecs), reverse=True))
X_pc1 = eigvecs # Edit

Thanks a lot to @Kirill ! He found my mistakes and the problem is solved now! Here is the correct version for future reference:

from sklearn.preprocessing import KernelCenterer
from scipy.spatial.distance import pdist, squareform
from scipy import exp
from scipy.linalg import eigh

# pdist to calculate the squared Euclidean distances for every pair of points
# in the 100x2 dimensional dataset.
sq_dists = pdist(X, 'sqeuclidean')

# Variance of the Euclidean distance between all pairs of data points.
variance = np.var(sq_dists)

# squareform to converts the pairwise distances into a symmetric 100x100 matrix
mat_sq_dists = squareform(sq_dists)

# set the gamma parameter equal to the one I used in scikit-learn KernelPCA
gamma = 15

# Compute the 100x100 kernel matrix
K = exp(-gamma * mat_sq_dists)

# Center the kernel matrix
kern_cent = KernelCenterer()
K = kern_cent.fit_transform(K)

# Get eigenvalues in ascending order with corresponding
# eigenvectors from the symmetric matrix
eigvals, eigvecs = eigh(K)

# Get the eigenvectors that corresponds to the highest eigenvalue
X_pc1 = eigvecs[:,-1] • Just a guess, and I haven't run your code, but in exp(gamma * mat_sq_dists), is the sign of gamma correct? Shouldn't it be $-15$ instead of $15$? – Kirill Sep 15 '14 at 0:10
• Thanks for the comment! I have actually tried both variants... The equation is written differently e.g., openclassroom.stanford.edu/MainFolder/… like you said, and the other way around on Wikipedia en.wikipedia.org/wiki/Radial_basis_function_kernel – user39663 Sep 15 '14 at 0:27
• I don't mean in the equation. In your code, if gamma is positive, then the kernel tends to infinity as two points tend away from each other, which is the opposite of what I would expect, i.e., $\exp(15\|x_1-x_2\|^2)\to\infty$ instead of $\to0$. – Kirill Sep 15 '14 at 0:30
• @Kirill is right, it should be minus gamma. – Marc Claesen Sep 15 '14 at 7:07
• @MarcClaesen Yes, he was right. I played with different lambdas (positive and negative) but the biggest problem was probably that I selected the first eigenvector wrongly. I selected the first row of the eigenvector matrix instead of the first column. Thanks to Kirill the problem is solved now :) Thanks everyone – user39663 Sep 15 '14 at 14:27

## 1 Answer

The first problem seems to be that the sign of gamma is wrong (it should be negative: $-15$, as in the definition of the kernel, not as in your code). Alternatively, use exp(-gamma * mat_sq_dists).

The second problem is that you clobber the eigenvectors with your invocation of zip's when you sort the list. The $i$-th eigenvector is eigvecs[:,i], not eigvecs[i,:], according to scipy.linalg.eigh (also: you should prefer eigh to eig because you have a symmetric real matrix).

Replace

< gamma = 15
> gamma = -15


and (to get ordered, real eigenvalues)

< eigvals, eigvecs = np.linalg.eig(K)
> eigvals, eigvecs = scipy.linalg.eigh(K)


and

< eigvals, eigvecs = zip(*sorted(zip(eigvals, eigvecs), reverse=True))
< X_pc1 = eigvecs
> X_pc1 = eigvecs[:,99]


Finally, you can examine scikit-learn's own implementation here.

• Thanks you so much! I really really appreciate it! It was such a frustrating problem and I am so happy that this is solved now thanks to you! I added the correct solution add the bottom of my OP. Thousand thanks! – user39663 Sep 15 '14 at 3:11
• @SebastianRaschka You're welcome. – Kirill Sep 15 '14 at 3:26